Prove by PMI that :- 1 × 1! + 2×2! + 3×3! + ... + n × n! = (n+1)! - 1

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{P}\left(\mathrm{n}\right): 1.1!+2.2!+...+\mathrm{n}.\mathrm{n}!=\left(\mathrm{n}+1\right)!-1 \\ \mathrm{Step} 1: \mathrm{Base} \mathrm{Case} \\ \mathrm{For} \mathrm{n}=1 \\ 1.1!=\left(1+1\right)!-1 \\ \Rightarrow 1=1 \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=1 \\ \mathrm{Step} 2: \mathrm{Inductive} \mathrm{Hypothesis} \\ \mathrm{Let} \mathrm{P}\left(\mathrm{n}\right) \mathrm{be} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k} \\ 1.1!+2.2!+...+\mathrm{k}.\mathrm{k}!=\left(\mathrm{k}+1\right)!-1...\left(\mathrm{i}\right) \\ \mathrm{Step} 3: \mathrm{Inductive} \mathrm{Step} \\ \mathrm{We} \mathrm{have} \\ 1.1!+2.2!+...+\mathrm{k}.\mathrm{k}!+\left(\mathrm{k}+1\right).\left(\mathrm{k}+1\right)! \\ \left(\mathrm{k}+1\right)!-1+\left(\mathrm{k}+1\right).\left(\mathrm{k}+1\right)! \left[\mathrm{By} \left(\mathrm{i}\right)\right] \\ =\left(\mathrm{k}+1\right)!+\left(\mathrm{k}+1\right).\left(\mathrm{k}+1\right)!-1 \\ =\left(\mathrm{k}+1\right)! \left(1+\mathrm{k}+1\right)-1 \\ =\left(\mathrm{k}+2\right)\left(\mathrm{k}+1\right)!-1 \\ =\left(\mathrm{k}+2\right)!-1 \\ \Rightarrow 1.1!+2.2!+...+\mathrm{k}.\mathrm{k}!+\left(\mathrm{k}+1\right).\left(\mathrm{k}+1\right)!=\left(\mathrm{k}+2\right)!-1 \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1 \\ \mathrm{So} \mathrm{by} \mathrm{principle} \mathrm{of} \mathrm{Mathematical} \mathrm{Induction}, \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{all} \mathrm{n}\in \mathrm{N} \end{gathered}$$

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